Print a Binary Tree in Vertical Order - All In One

LeetCode 314 Binary Tree Vertical Order Traversal
Print a Binary Tree in Vertical Order | Set 1 | GeeksforGeeks
Given a binary tree, print it vertically. The following example illustrates vertical order traversal
The idea is to traverse the tree once and get the minimum and maximum horizontal distance with respect to root. 
Once we have maximum and minimum distances from root, we iterate for each vertical line at distance minimum to maximum from root, and for each vertical line traverse the tree and print the nodes which lie on that vertical line.

void findMinMax(Node *node, int *min, int *max, int hd)

{

    // Base case

    if (node == NULL) return;

    // Update min and max

    if (hd < *min)  *min = hd;

    else if (hd > *max) *max = hd;

    // Recur for left and right subtrees

    findMinMax(node->left, min, max, hd-1);

    findMinMax(node->right, min, max, hd+1);

}

// The main function that prints a given binary tree in

// vertical order

void verticalOrder(Node *root)

{

    // Find min and max distances with resepect to root

    int min = 0, max = 0;

    findMinMax(root, &min, &max, 0);

    // Iterate through all possible vertical lines starting

    // from the leftmost line and print nodes line by line

    for (int line_no = min; line_no <= max; line_no++)

    {

        printVerticalLine(root, line_no, 0);

        cout << endl;

    }

}

Time Complexity: Time complexity of above algorithm is O(w*n) where w is width of Binary Tree and n is number of nodes in Binary Tree. In worst case, the value of w can be O(n) (consider a complete tree for example) and time complexity can become O(n²).
http://buttercola.blogspot.com/2014/12/facebook-print-binary-tree-in-vertical.html

O(N): Print a Binary Tree in Vertical Order | Set 2 (Hashmap based Method) | GeeksforGeeks
We need to check the Horizontal Distances from root for all nodes. If two nodes have the same Horizontal Distance (HD), then they are on same vertical line. The idea of HD is simple. HD for root is 0, a right edge (edge connecting to right subtree) is considered as +1 horizontal distance and a left edge is considered as -1 horizontal distance.

We can do inorder traversal of the given Binary Tree. While traversing the tree, we can recursively calculate HDs. We initially pass the horizontal distance as 0 for root. For left subtree, we pass the Horizontal Distance as Horizontal distance of root minus 1. For right subtree, we pass the Horizontal Distance as Horizontal Distance of root plus 1. For every HD value, we maintain a list of nodes in a hasp map. Whenever we see a node in traversal, we go to the hash map entry and add the node to the hash map using HD as a key in map.

void getVerticalOrder(Node* root, int hd, map<int, vector<int>> &m)

{

    // Base case

    if (root == NULL)

        return;

    // Store current node in map 'm'

    m[hd].push_back(root->key);

    // Store nodes in left subtree

    getVerticalOrder(root->left, hd-1, m);

    // Store nodes in right subtree

    getVerticalOrder(root->right, hd+1, m);

}

// The main function to print vertical oder of a binary tree

// with given root

void printVerticalOrder(Node* root)

{

    // Create a map and store vertical oder in map using

    // function getVerticalOrder()

    map < int,vector<int> > m;

    int hd = 0;

    getVerticalOrder(root, hd,m);

    // Traverse the map and print nodes at every horigontal

    // distance (hd)

    map< int,vector<int> > :: iterator it;

    for (it=m.begin(); it!=m.end(); it++)

    {

        for (int i=0; i<it->second.size(); ++i)

            cout << it->second[i] << " ";

        cout << endl;

    }

}

Java Code: http://www.shuatiblog.com/blog/2014/12/17/Print-Binary-Tree-Vertically/

public List<List<Integer>> printVertically(TreeNode root) {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();

    // 1. find the range of left bound and right bound
    int[] range = new int[2];
    findRange(root, range, 0);

    // 2. calculate number of columns in the result
    int rootIndex = 0 - range[0];
    int columns = range[1] - range[0] + 1;
    for (int i = 0; i < columns; i++) {
        ans.add(new ArrayList<Integer>());
    }

    // 3. fill in vertically in a recursive manner
    fillNode(ans, root, rootIndex);

    return ans;
}

private void fillNode(List<List<Integer>> ans, TreeNode node, int index) {
    if (node == null) {
        return;
    }
    ans.get(index).add(node.val);
    fillNode(ans, node.left, index - 1);
    fillNode(ans, node.right, index + 1);
}

private void findRange(TreeNode node, int[] range, int position) {
    if (node == null) {
        return;
    }
    if (position < range[0]) {
        range[0] = position;
    }
    if (position > range[1]) {
        range[1] = position;
    }
    findRange(node.left, range, position - 1);
    findRange(node.right, range, position + 1);
}

http://algorithms.tutorialhorizon.com/print-the-binary-tree-in-vertical-order-path/

• Do the inorder traversal.
• Take a vari­able called level, when ever you go left, do level++ AND when ever you go right do level–.
• With step above we have sep­a­rated out the lev­els vertically.
• Now you need to store the ele­ments of each level, so cre­ate a TreeMap and the (key,value) pair will be (level,elements at that level).
• At the end iter­ate through the TreeMap and print the results.

http://codecrazy-me.blogspot.com/2014/09/printing-vertical-level-nodes-of-binary.html
http://algorithms.tutorialhorizon.com/print-the-binary-tree-in-vertical-order-path/
Solution 3: Use Level Order:
http://yuanhsh.iteye.com/blog/2219614

但是上面两种方法都不能保证从上往下看的顺序。所以要保证从上到下的顺序的话，需要用BFS即Level Order的方法来遍历。用pair<int, TreeNode*>来保存index和节点指针。

1. vector<vector<int>> vertical_traversal(TreeNode* root) {  
2.     vector<vector<int>> res;  
3.     if(!root) return res;  
4.     unordered_map<int, vector<int>> map;  
5.     queue<pair<int, TreeNode*>> queue;  
6.     queue.emplace(0, root);  
7.     int left=0, right=0;  
8.     while(!queue.empty()) {  
9.         auto& p = queue.front();  
10.         queue.pop();  
11.         left = min(left, p.first);  
12.         right = max(right, p.first);  
13.         map[p.first].push_back(p.second->val);  
14.         if(p.second->left)   
15.             queue.emplace(p.first-1, p.second->left);  
16.         if(p.second->right)   
17.             queue.emplace(p.first+1, p.second->right);  
18.     }  
19.     for(int i=left; i<=right; i++)   
20.         res.push_back(map[i]);  
21.     return res;  
22. } 

http://javabypatel.blogspot.com/2015/10/print-binary-tree-in-vertical-order.html
In above solution, Vertical order traversal of a tree is working fine, but nodes at same level are not printed in sequence they are present because we are doing Pre-order traversal of a tree (Depth first traversal) in which Left sub-tree nodes are read first and then it picks right sub-tree.

For mainlining sequence, we have to do Breadth first traversal that is Level order traversal, which make sure nodes are Lower levels are read first than Nodes at higher level that is all Nodes at Level 0 are read first and then Node at Level 1 and then Node at Level 2 and so on.

For this we have to take a queue, which will preserve nodes at same level for later processing.

private static void printTreeInVerticalOrderMaintainSequence(VerticalOrderNode rootNode, TreeMap<Integer, String> map) {

 if(rootNode==null){

  return;

 }

 Queue<VerticalOrderNode> q = new LinkedList<VerticalOrderNode>();

 q.add(rootNode);

 while(!q.isEmpty()){

  VerticalOrderNode temp = q.poll();

  if(map.get(temp.getLevel())!=null){

   map.put(temp.getLevel(), map.get(temp.getLevel())+","+temp.getData());

  }else{

   map.put(temp.getLevel(), String.valueOf(temp.getData()));

  }

  if(temp.getLeftNode()!=null){

   temp.getLeftNode().setLevel(temp.getLevel()-1);

   q.add(temp.getLeftNode());

  }

  if(temp.getRightNode()!=null){

   temp.getRightNode().setLevel(temp.getLevel()+1);

   q.add(temp.getRightNode());

  }

 }

}

class VerticalOrderNode{

 private int data;

 private VerticalOrderNode leftNode;

 private VerticalOrderNode rightNode;

 private int level; // don't have to

}

http://www.ideserve.co.in/learn/print-binary-tree-vertical-order

12	class QueueEntry
13	{
14	TreeNode node;
15	int horizontalDistance;
16	QueueEntry(TreeNode node, int horizontalDistance)
17	{
18	this.node = node;
19	this.horizontalDistance = horizontalDistance;
20	}
21	}
35	TreeNode root;
36
37	private void fillUpVerticalOrderMap(TreeNode currentNode, int horizontalDistFromRoot, Map verticalOrderMap)
38	{
39	if (currentNode == null) return;
40
41	ArrayList<Integer> mapEntry;
42	LinkedList<QueueEntry> queue = new LinkedList();
43
44	queue.add(new QueueEntry(currentNode, horizontalDistFromRoot));
45
46	while (!queue.isEmpty())
47	{
48	QueueEntry entry = queue.remove();
49
50
51	mapEntry = (ArrayList<Integer>) verticalOrderMap.get(entry.horizontalDistance);
52
53	if (mapEntry != null)
54	{
55	mapEntry.add(entry.node.val);
56	verticalOrderMap.put(entry.horizontalDistance, mapEntry);
57	}
58	else
59	{
60	mapEntry = new ArrayList();
61	mapEntry.add(entry.node.val);
62	verticalOrderMap.put(entry.horizontalDistance, mapEntry);
63	}
64
65
66	if (entry.node.left != null)
67	{
68
69	queue.add(new QueueEntry(entry.node.left, entry.horizontalDistance - 1));
70	}
71
72	if (entry.node.right != null)
73	{
74
75	queue.add(new QueueEntry(entry.node.right, entry.horizontalDistance + 1));
76	}
77	}
78	}
79
80
81	public void printVerticalOrder()
82	{
83	Map<Integer, ArrayList<Integer>> verticalOrderMap = new TreeMap<>();
84
85	fillUpVerticalOrderMap(root, 0, verticalOrderMap);
86
87
88	Iterator<Entry<Integer, ArrayList<Integer>>> iterator = verticalOrderMap.entrySet().iterator();
89
90	while (iterator.hasNext())
91	{
92	Entry<Integer, ArrayList<Integer>> mapEntry = iterator.next();
93	ArrayList<Integer> nodeList = mapEntry.getValue();
94	for (int i = 0; i < nodeList.size(); i++)
95	{
96	System.out.print("  "  + nodeList.get(i));
97	}
98	System.out.println("" );
99	}
100	}

https://shawnlincoding.wordpress.com/2015/04/12/facebook-meetqun/
follow up要one pass (不能找最左有多远，最右有多远)， 不要hashmap
把root作为中心，index为0，向左移动-1，向右移动+1，把相同level的点存在同一个list里
    public void print(TreeNode root) {
        ArrayList<ArrayList<Integer>> right = new ArrayList<ArrayList<Integer>>();
        ArrayList<ArrayList<Integer>> left = new ArrayList<ArrayList<Integer>>();
        printHelper(0, root, left, right);
        // Print result
        for (int i = left.size() - 1; i > 0; i--) {
            System.out.println(left.get(i));
        }
        for (int i = 0; i < right.size(); i++) {
            System.out.println(right.get(i));
        }
    }

    private void printHelper(int index, TreeNode root, ArrayList<ArrayList<Integer>> left, ArrayList<ArrayList<Integer>> right) {
        // Base case
        if (root == null) {
            return;
        }
        // Normal case
        if (index >= 0) {
            while (right.size() <= index) {
                right.add(new ArrayList<Integer>());
            }
            right.get(index).add(root.val);
        } else {
            while (left.size() <= -index) {    // 注意这边用的是while, 因为index=0的时候是把点加到right里，所以left在index=0的时候得加一个空List
                left.add(new ArrayList<Integer>());
            }
            left.get(-index).add(root.val);
        }
        // Recurse
        printHelper(index - 1, root.left, left, right);
        printHelper(index + 1, root.right, left, right);
    }

Find Vertical Sum of given Binary Tree

---

Given a Binary Tree, find vertical sum of the nodes that are in same vertical line.
Print all sums through different vertical lines.
Let us first understand what we want to achieve? what is the input and what will be the expected output?
If we change the logic which concat the data present at same vertical line to add data at same vertical line, then our task is done. 

Uber Prepare: Print Binary Tree With No Two Nodes Share The Same Column
http://www.cnblogs.com/EdwardLiu/p/6367347.html

Give a binary tree, elegantly print it so that no two tree nodes share the same column. 

Requirement: left child should appear on the left column of root, and right child should appear on the right of root.

Example: 
    a
   b   c
 d   e   f
z g   h i j

这道题若能发现inorder traversal each node的顺序其实就是column number递增的顺序，那么就成功了一大半

维护一个global variable，colNum, 做inorder traversal

然后level order 一层一层打印出来

 6     public static class TreeNode {
 7         char val;
 8         int col;
 9         TreeNode left;
10         TreeNode right;
11         public TreeNode(char value) {
12             this.val = value;
13         }
14     }
15     
16     static int colNum = 0;
17     
18     public static List<String> print(TreeNode root) {
19         List<String> res = new ArrayList<String>();
20         if (root == null) return res;
21         inorder(root);
22         levelOrder(root, res);
23         return res;
24     }
25     
26     public static void inorder(TreeNode node) {
27         if (node == null) return;
28         inorder(node.left);
29         node.col = colNum;
30         colNum++;
31         inorder(node.right);
32         
33     }
34     
35     public static void levelOrder(TreeNode node, List<String> res) {
36         Queue<TreeNode> queue = new LinkedList<TreeNode>();
37         queue.offer(node);
38         while (!queue.isEmpty()) {
39             StringBuilder line = new StringBuilder();
40             HashMap<Integer, Character> lineMap = new HashMap<Integer, Character>();
41             int maxCol = Integer.MIN_VALUE;
42             int size = queue.size();
43             for (int i=0; i<size; i++) {
44                 TreeNode cur = queue.poll();
45                 lineMap.put(cur.col, cur.val);
46                 maxCol = Math.max(maxCol, cur.col);
47                 if (cur.left != null) queue.offer(cur.left);
48                 if (cur.right != null) queue.offer(cur.right);
49             }
50             for (int k=0; k<=maxCol; k++) {
51                 if (lineMap.containsKey(k)) line.append(lineMap.get(k));
52                 else line.append(' ');
53             }
54             res.add(line.toString());
55         }
56     }

89         List<String> res = print(A);
90         for (String each : res) {
91             System.out.println(each);
92         }

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